67.THE RATIO OF E2-E1 AND E4-E3 FOR THE HYDROGEN ATOM IS APPROXIMATELY EQUAL TO
(A)10
(B)15
(C)17
(D)12.

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Solution

The explanation for the correct option
Option(B) 15
for the hydrogen atom, the energy is given as:
$E = - 13.6 \frac{z^{2}}{n^{2}} w h e r e, z = c h a r g e, n = n u m b e r o f o r b i t$

The difference in energy is given as:
$Δ E = E_{0} z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
The ratio of $E_{2} - E_{1} a n d E_{4} - E_{3}$ for the hydrogen atom can be calculated as:
$∴ E_{2} - E_{1} = E_{0} 1^{2} (\frac{1}{1} - \frac{1}{4}) = \frac{3}{4} E_{0}$
$E_{4} - E_{3} = E_{0} 1^{2} (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} E_{0}$
$∴ R a t i o = \frac{E_{2} - E_{1}}{E_{4} - E_{3}} = \frac{\frac{3}{4} E_{0}}{\frac{7}{144} E_{0}} = \frac{3 \times 144}{4 \times 7} = 15.4 \approx 15$
Hence, Option(B) 15 is the ratio of
$E_{2} - E_{1} a n d E_{4} - E_{3}$
for the hydrogen atom.

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